Problem 30 - A Graphical Approach to Precalculus with Limits

Question

Solve \((\mathbf{a}) f(x)=0,(\mathbf{b}) f(x)>0,\) and \((\mathbf{c}) f(x)<0\) over the interval \([0,2 \pi)\) $$f(x)=2 \sin ^{2} x+3 \sin x+1$$

Step-by-Step Solution

Answer

\(f(x) = 0\) at \(x = \frac{3\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}\); \(f(x) > 0\) on \((0, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi)\); \(f(x) < 0\) on \((\frac{7\pi}{6}, \frac{11\pi}{6})\).

Step 1: Understanding the Function

First, we have the function \(f(x) = 2 \sin^2 x + 3 \sin x + 1\). This function is quadratic in terms of \(\sin x\). It is helpful to think of it as \(f(\sin x) = 2u^2 + 3u + 1\) by letting \(u = \sin x\).

Step 2: Solving for \(f(x) = 0\)

To solve \(f(x) = 0\), set \(2 \sin^2 x + 3 \sin x + 1 = 0\) and solve for \(\sin x\). This is equivalent to solving the quadratic equation \(2u^2 + 3u + 1 = 0\). The solutions are found using the quadratic formula: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For \(a = 2\), \(b = 3\), and \(c = 1\), the solutions are \(u = \sin x = -1\) and \(u = \sin x = -\frac{1}{2}\). Hence, \(x = \frac{3\pi}{2}\) for \(\sin x = -1\), and \(x = \frac{7\pi}{6}, \frac{11\pi}{6}\) for \(\sin x = -\frac{1}{2}\).

Step 3: Solving for \(f(x) > 0\)

To solve \(f(x) > 0\), find intervals where \(2 \sin^2 x + 3 \sin x + 1 > 0\). Knowing the roots from step 2, determine intervals between these values: when moving from \(x = 0\) to \(x = \frac{7\pi}{6}\), then \(x = \frac{7\pi}{6}\) to \(x = \frac{11\pi}{6}\), and finally \(x = \frac{11\pi}{6}\) to \(x = 2\pi\). Test points show: 1. \(x = \frac{\pi}{2}\) gives \(f(x) > 0\).2. \(x = \frac{3\pi}{2}\) gives \(f(x) > 0\).Thus, the intervals where \(f(x) > 0\) are \( (0, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi) \).

Step 4: Solving for \(f(x) < 0\)

To solve \(f(x) < 0\), find intervals where \(2 \sin^2 x + 3 \sin x + 1 < 0\). As established, these intervals are between the roots \(\frac{7\pi}{6}\) and \(\frac{11\pi}{6}\). Test points such as \(x = \pi\) show \(f(x) < 0\), confirming this interval. Thus, the interval is \( (\frac{7\pi}{6}, \frac{11\pi}{6}) \).

Key Concepts

Quadratic Functions

At the core of many mathematical problems lies the quadratic function. A quadratic function is an equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients, and \(a eq 0\). It forms a parabola when graphed. In our given exercise, we encountered a quadratic equation in terms of \(\sin x\), represented as \(2\sin^2 x + 3\sin x + 1\). By letting \(u = \sin x\), the equation transforms to \(2u^2 + 3u + 1\). This is a typical quadratic equation where:

• \(a = 2\)
• \(b = 3\)
• \(c = 1\)

To find the roots or solutions for \(f(x) = 0\), we use the quadratic formula:\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula calculates the values of \(u\) that satisfy the equation. By substituting the values of \(a\), \(b\), and \(c\) into the formula, we find the roots that help in solving the trigonometric equation \(\sin x = u\).

Interval Notation

Interval notation is a shorthand used to describe sets of numbers within an interval on a number line. This is particularly useful in solving inequalities, such as when we have established conditions like \(f(x) > 0\). In the given exercise, interval notation helps us precisely describe the regions on the graph of \(x\) where the equation satisfies certain conditions, such as being greater than or less than zero.To solve quadratic inequalities in the context of our trigonometric function, we determine the roots (or solving for the values where the equation equals zero). The solutions and the resulting intervals are expressed using bracket notation:

• \([a, b]\) – Including \(a\) and \(b\), known as a closed interval.
• \((a, b)\) – Excluding \(a\) and \(b\), known as an open interval.

The exercise determined intervals of positivity and negativity through test points in regions defined by the roots found during solving of \(f(x) = 0\). For instance, the positive intervals \((0, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi)\) indicate the regions where the function value is greater than zero.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the involved variables. They are essential in simplifying and solving trigonometric equations like the one in this exercise. Understanding these identities allows us to manipulate and transform equations into solvable forms.For example, the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\) might be useful in other trigonometric contexts to replace \(\sin^2 x\) or \(\cos^2 x\) accordingly. By knowing and applying trigonometric identities, we can sometimes simplify our original functions into recognizable patterns that lay the groundwork for using methods like substitution, as seen with \(\sin x = u\) in our exercise.In the given problem, recognizing that \(f(x)\) can be viewed as a quadratic in terms of \(\sin x\) was key. Although direct trigonometric identities weren't transformed in this step, the concept aids in understanding how we substitute and solve for \(\sin x\) to find when it equals specific values based on identity-based adjustments.